Problem: The surface area of a cylinder is increasing at a rate of $9\pi$ square meters per hour. The height of the cylinder is fixed at $3$ meters. At a certain instant, the surface area is $36\pi$ square meters. What is the rate of change of the volume of the cylinder at that instant (in cubic meters per hour)? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\pi}{3}$ (Choice B) B $27\pi$ (Choice C) C $9\pi$ (Choice D) D $\dfrac{1}{2}$ The surface area of a cylinder with base radius $r$ and height $h$ is $2\pi r^2+2\pi r h$. The volume of a cylinder with base radius $r$ and height $h$ is $\pi r^2h$.
Solution: Setting up the math Let... $r(t)$ denote the cylinder's base radius at time $t$, $h$ denote the cylinder's height (which is always $3$ meters), $V(t)$ denote the cylinder's volume at time $t$, and $S(t)$ denote the cylinder's surface area at time $t$. We are given that $h=3$ and $S'(t)=9\pi$, We are also given that $S(t_0)=36\pi$ for a specific time $t_0$. We want to find $V'(t_0)$. Relating the measures $S(t)$ and $r(t)$ relate to each other through the formula for the surface area of a cylinder: $\begin{aligned} S(t)&=2\pi [r(t)]^2+2\pi r(t)h \\\\ &=2\pi[r(t)]^2+6\pi r(t) \end{aligned}$ We can differentiate both sides to find an expression for $S'(t)$ : $S'(t)=4\pi r(t)r'(t)+6\pi r'(t)$ $V(t)$ and $r(t)$ relate to each other through the formula for the volume of a cylinder: $\begin{aligned} V(t)&=\pi[r(t)]^2h \\\\ &=3\pi[r(t)]^2 \end{aligned}$ We can differentiate both sides to find an expression for $V'(t)$ : $V'(t)=6\pi r(t)r'(t)$ Using the information to solve Let's plug ${S(t_0)}={36\pi}$ into the expression for $S(t_0)$ : $\begin{aligned} {S(t_0)}&=2\pi[r(t_0)]^2+6\pi r(t_0) \\\\ {36\pi}&=2\pi[r(t_0)]^2+6\pi r(t_0) \\\\ 18&=[r(t_0)]^2+3r(t_0) \\\\ 0&=[r(t_0)]^2+3r(t_0)-18 \\\\ 0&=[r(t_0)+6][r(t_0)-3] \\\\ {3}&={r(t_0)} \end{aligned}$ Let's plug ${S'(t_0)}={9\pi}$ and ${r(t_0)}={3}$ into the expression for $S'(t_0)$ : $\begin{aligned} {S'(t_0)}&=4\pi{r(t_0)}r'(t)+6\pi r'(t_0) \\\\ {9\pi}&=4\pi( 3)r'(t_0)+6\pi r'(t_0) \\\\ 9&=18 r'(t) \\\\ C{\dfrac12}&=C{r'(t_0)} \end{aligned}$ Now let's plug ${r(t_0)}={3}$ and $C{r'(t_0)}=C{\dfrac{1}{2}}$ into the expression for $V'(t_0)$ : $\begin{aligned} V'(t_0)&=6\pi{r(t_0)}C{r'(t_0)} \\\\ &=6\pi({3})\left(C{\dfrac{1}{2}}\right) \\\\ &=9\pi \end{aligned}$ In conclusion, the rate of change of the volume of the cylinder at that instant is $9\pi$ cubic meters per hour. Since the rate of change is positive, we know that the volume is increasing.